Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{a^2 - 16a + 63}{10a + 70} \div \dfrac{a^2 - 9a}{-4a - 28} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{a^2 - 16a + 63}{10a + 70} \times \dfrac{-4a - 28}{a^2 - 9a} $ First factor the quadratic. $q = \dfrac{(a - 9)(a - 7)}{10a + 70} \times \dfrac{-4a - 28}{a^2 - 9a} $ Then factor out any other terms. $q = \dfrac{(a - 9)(a - 7)}{10(a + 7)} \times \dfrac{-4(a + 7)}{a(a - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (a - 9)(a - 7) \times -4(a + 7) } { 10(a + 7) \times a(a - 9) } $ $q = \dfrac{ -4(a - 9)(a - 7)(a + 7)}{ 10a(a + 7)(a - 9)} $ Notice that $(a + 7)$ and $(a - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -4\cancel{(a - 9)}(a - 7)(a + 7)}{ 10a(a + 7)\cancel{(a - 9)}} $ We are dividing by $a - 9$ , so $a - 9 \neq 0$ Therefore, $a \neq 9$ $q = \dfrac{ -4\cancel{(a - 9)}(a - 7)\cancel{(a + 7)}}{ 10a\cancel{(a + 7)}\cancel{(a - 9)}} $ We are dividing by $a + 7$ , so $a + 7 \neq 0$ Therefore, $a \neq -7$ $q = \dfrac{-4(a - 7)}{10a} $ $q = \dfrac{-2(a - 7)}{5a} ; \space a \neq 9 ; \space a \neq -7 $